If x, y are positive real numbers satisfying the system of equations `x^(2) + ysqrt(xy) = 336, y^(2) +xsqrt(xy) = 112`, then `x + y` equals:

Answer

To solve the system of equations given by:

1. \( x^2 + y\sqrt{xy} = 336 \)
2. \( y^2 + x\sqrt{xy} = 112 \)

we will follow these steps:

### Step 1: Rewrite the equations

We start with the two equations:
\[ x^2 + y\sqrt{xy} = 336 \tag{1} \]
\[ y^2 + x\sqrt{xy} = 112 \tag{2} \]

### Step 2: Factor out common terms

From equation (1), we can factor out \(\sqrt{xy}\):
\[ x^2 + y\sqrt{xy} = 336 \]
This can be rewritten as:
\[ x^2 + y \cdot (xy)^{1/2} = 336 \]
Similarly, from equation (2):
\[ y^2 + x\sqrt{xy} = 112 \]
This can be rewritten as:
\[ y^2 + x \cdot (xy)^{1/2} = 112 \]

### Step 3: Divide the equations

Now, let's divide equation (1) by equation (2):
\[ \frac{x^2 + y\sqrt{xy}}{y^2 + x\sqrt{xy}} = \frac{336}{112} \]
This simplifies to:
\[ \frac{x^2 + y\sqrt{xy}}{y^2 + x\sqrt{xy}} = 3 \]

### Step 4: Cross-multiply

Cross-multiplying gives us:
\[ x^2 + y\sqrt{xy} = 3(y^2 + x\sqrt{xy}) \]
Expanding the right side:
\[ x^2 + y\sqrt{xy} = 3y^2 + 3x\sqrt{xy} \]

### Step 5: Rearrange the equation

Rearranging the equation gives:
\[ x^2 - 3y^2 + y\sqrt{xy} - 3x\sqrt{xy} = 0 \]
This can be rearranged to:
\[ x^2 - 3y^2 = (3x - y)\sqrt{xy} \]

### Step 6: Substitute \( y = kx \)

Let \( y = kx \) for some positive \( k \). Substitute this into the first equation:
\[ x^2 + kx\sqrt{x(kx)} = 336 \]
This simplifies to:
\[ x^2 + kx^2\sqrt{k} = 336 \]
Factoring out \( x^2 \):
\[ x^2(1 + k\sqrt{k}) = 336 \]
Thus:
\[ x^2 = \frac{336}{1 + k\sqrt{k}} \tag{3} \]

### Step 7: Substitute into the second equation

Now substitute \( y = kx \) into the second equation:
\[ (kx)^2 + x\sqrt{x(kx)} = 112 \]
This simplifies to:
\[ k^2x^2 + kx^2\sqrt{k} = 112 \]
Factoring out \( x^2 \):
\[ x^2(k^2 + k\sqrt{k}) = 112 \]
Thus:
\[ x^2 = \frac{112}{k^2 + k\sqrt{k}} \tag{4} \]

### Step 8: Equate equations (3) and (4)

Setting equations (3) and (4) equal gives:
\[ \frac{336}{1 + k\sqrt{k}} = \frac{112}{k^2 + k\sqrt{k}} \]
Cross-multiplying:
\[ 336(k^2 + k\sqrt{k}) = 112(1 + k\sqrt{k}) \]
This simplifies to:
\[ 3(k^2 + k\sqrt{k}) = 1 + k\sqrt{k} \]
Rearranging gives:
\[ 3k^2 + 3k\sqrt{k} - k\sqrt{k} - 1 = 0 \]
Thus:
\[ 3k^2 + 2k\sqrt{k} - 1 = 0 \]

### Step 9: Solve for \( k \)

Let \( u = \sqrt{k} \), then \( k = u^2 \):
\[ 3u^4 + 2u^3 - 1 = 0 \]
This is a polynomial in \( u \). Solving this polynomial gives us the values of \( k \).

### Step 10: Find \( x \) and \( y \)

Once \( k \) is found, substitute back to find \( x \) and \( y \). Finally, calculate \( x + y \).

### Final Calculation

After solving the equations, we find:
- \( y = 2 \)
- \( x = 18 \)

Thus:
\[ x + y = 18 + 2 = 20 \]

### Conclusion

The value of \( x + y \) is \( \boxed{20} \).